By Corollary REF , \(m_n \le n^2+1\) , and in particular \(m_2\le 5\) . Even the answer \(m_2 = 5\)  given in [1]},
is not easy, although a reasonably fast combinatoric approach is
given in [2]}, where it is observed that any since any
Lie algebra \(L\)  satisfying a Lie identity of degree \(<5\)  is
solvable, one must have \(m_2 \ge 5\) , yielding \(m_2 = 5\) .
Špenko [3]} looked at this from the other
direction and showed that if \(p\)  is a Lie polynomial of degree
\(\le 4\)  then \(\operatorname{Im}p=\operatorname{sl}_2\) .
